By Sohan Crasta
25/05/2015

Equilibrium and Solubility

• All equilibrium equations are dissolving equations (i.e. a solid splitting into its ions)
• Keq changes to ksp, where "sp" stands for solubility product
• Ksp indirectly represents the amount of solute that can dissolve
• Ksp also represents the max concentration of the ions at equilibrium (i.e. saturation)

Relating Ksp to Solubility of a Salt

• Ksp is dimensionless but is an exponent of mol/L
• Solubility is usually measured in g/100mL or g/L
• the "x" value from the Ksp expression represents the solubility of the salt in mol/L (i.e. molar solubility)

Video on finding Ksp

Homework:

Solubility Problems worksheet questions
Study for Acid-Base Equilibrium test tomorrow

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By Hafshan Hameem
28/05/2015

Predicting Precipitation

- In order to determine if a precipitate will form or not (or if total amount of a solute will dissolve) you must determine the TRIAL ION PRODUCT (Q)
- Q is found by substituting the "ACTUAL ION CONCENTRATIONS" into the Ksp expression and comparing that value with the Ksp
- If Q> Ksp a precipitate will form ( or not all will dissolve
- If Q < or = a precipitate will not form ( or all will dissolve)

Sample Problem

If 425 mL of 0.15 M calcium chloride reacts with 450 mL of 0.25 silver nitrate
Determine A) If a precipitate will form

A) CaCl2 (aq) + 2AgNO3(aq) ---------> Ca(NO3)2 (aq) + 2AgCl(s)
c 0.15 0.25
v 0.425 0.450 ---------> Volume total = 0.875 L
n

AgCl(s) ⇄ Ag+ (aq) + Cl- (aq)

Ksp= [Ag+][Cl-]= 1.8x10-10


C1V1=C2V2

[Ag+] actual= (0.25M)(0.45L)/ 0.875 L
= 0.13 M

[Cl-] actual= (2)(0.15M)(0.425 L)/ 0.875 L
= 0.15 M

Q= [Ag+][Cl-]
= (0.13)(0.15 M)
Q= 2.0x10-2
Q> Ksp
Therefore a precipitate (ppt) will form